Question

What is the minimum amount of time required to measure the kinetic energy of an electron with a speed 100 +- 1 m/s.

Answer 1

u=600msec−1

∴Δu=600×0.005100

⇒0.30msec−1

Now,Δx.m.Δu=h4π

Δx=6.6×10−344×3.14×9.1×10−31×0.30

⇒1.92×10−3m

Answer 2

ΔE.Δt ≥ h/4π

v=(100 ± 1) m/s

ΔE= 1/2 mΔv²

     = 1/2 * 9.1 * 10^-31 *((101)²-(99)²)

     = 1.82*10^-28 J

Δt ≥ h/(4π*ΔE)

Δt ≥ 2.89 * 10^-7 seconds

minimum time= 2.89 * 10^-7 seconds

                       = 0.289 μs = 0.3 μs